Admissibility of groups over function fields of p-adic curves

0.18 MB·English

by  B. Surendranath Reddy

Admissibility of groups over function fields of
p-adic curves
2 B. Surendranath Reddy and V. Suresh
1
0
2

n Abstract
a
J

9 Let K be a field and G a finite group. The question of ‘admissibility’ of G

over K was originally posed by Schacher, who gave partial results in the case
]
A
K = Q. In this paper, we give necessary conditions for admissibility of a
R
finite group G over function fields of curves over complete discretely valued
.
h fields. Using this criterion, we give an example of a finite group which is not
t
a admissible over Q (t). We also prove a certain Hasse principle for division
p
m
algebras over such fields.
[

1
v Introduction
8
3
9 Let K be a field and G a finite group. We say that G is admissible over K if
1
there exists a division ring D central over K and a maximal subfield L of D
.
1 which is Galois over K with Galois group G. Schacher asked, given a field K,
0
2 which finite groups are admissible over K and proved that if a finite group
1 G is admissible over Q, then every Sylow subgroup of G is metacyclic ([Sc],
:
v 4.1). This led to the conjecture that a finite group G is admissible over Q
i
X if and only if every Sylow subgroup of G is meta-cyclic. This conjecture has
r been proved for all solvable groups ([So]) and for certain non-solvable groups
a
of small order ([CS], [FS], [Fe1], [Fe2], [Fe3]).
Recently Harbater, Hartman and Krashen ([HHK2], 4.5) gave a charac-
terization of admissible groups over function fields of curves over complete
discretely valued fields with algebraically closed residue fields. In this paper,
we consider the function fields of curves over complete discretely valued fields
without any assumptions on the residue fields and prove the following
Theorem 1. Let K be a complete discretely valued field with residue field
k and F = K(X) be the function field of a curve X over K. Let G be a finite
1
group. Suppose that the order of G is coprime to char(k). If G is admissible
over F then every Sylow subgroup P of G has a normal series P P P
1 2
⊇ ⊇
such that
(1) P/P and P are cyclic
1 2
(2) P /P is admissible over a finite extension of the residue field of a
1 2
discrete valuation of F.
A main ingredient for the proof of the above theorem is the following
Hasse principle for central simple algebras, which has independent interest.
Theorem 2. Let K be a complete discretely valued field with residue field
k and F = K(X) be the function field of a curve X over K. Let A be a
central simple algebra over F of degree n = ℓr for some prime ℓ and r 1.
≥
Assume that ℓ is not equal to char(k) and K contains a primitive nth root of
unity. Then index(A) = index(A F ) for some discrete valuation v of F.
v
⊗
For the proof of the above theorem, we use the patching techniques of
([HHK1]). A similar Hasse principle is proved for quadratic forms over such
fields in ([CTPS], 3.1). In ([HHK3], 9.12), it is proved that if a central simple
algebra A over F (F as above), is split over F for all discrete valuations on
ν
F, then A is split over F.
There are some examples of classes of finite groups which are admissible
over the rational function fields. However there was no example, in the
literature, ofafinitegroupwhichisnotadmissibleoverQ (t). UsingTheorem
p
1, we giveanexample ofafinitegroupwhich isnotadmissible over Q (t). We
p
also prove admissibility of a certain class of groups over Q (t) using patching
p
techniques.
Theorem 3. Let K be a p-adic field and F the function field of a curve
over F. Let G be a finite group with order coprime to char(k). If every
Sylow subgroup of G is a quotient of Z4, then G is admissible over F.
In[FSS],itwasprovedthateveryabeliangrouponthreeorlessgenerators
is admissible over Q(t). We conclude by showing that every abelian group of
order n with four or less generators is admissible over Q(ζ)(t), where ζ is a
primitive nth root of unity.
1. Some Preliminaries
In this section we recall a few basic definitions and facts about division
algebras and patching techniques ([GS], [HH], [HHK1], [P], [S2], [Sc], [Sch],
2
[Ser]).
LetK beafieldandBr(K)betheBrauergroupofcentralsimplealgebras
over K. For an integer n 2, let Br(K) denote the n-torsion subgroup
n
≥
of Br(K). If A and B are two central simple algebras over K, we write
A B if A and B are isomorphic as K-algebras and we write A = B if
≃
they represent the same element in Br(K). Let n be an integer coprime to
char(K). Suppose E/K is a cyclic extension of degree n and σ a generator
∗
of Gal(E/K). For b K , let (E/K,σ,b) (or simply (E,σ,b)) be the K-
∈
algebra generated by E and y with yn = b and λy = yσ(λ) for all λ E.
∈
Then (E,σ,b) is a central simple algebra over K and represents an element
in Br(K). Suppose that K contains a primitive nth root of unity. Then
n
E = K(√n a) for some a K∗ and σ(√n a) = ζ√n a for a primitive nth root of
unity ζ K∗. The algeb∈ra (E,σ,b) is generated by x,y with xn = a, yn = b
∈
and xy = ζyx and we denote the algebra (E,σ,b) by (a,b) .
n
Suppose that ν is a discrete valuation of K with residue field k. Let n
be a natural number which is coprime to char(k). Then we have a residue
homomorphism ∂ : Br(K) H1(k,Z/nZ), where H1(k,Z/nZ) denotes
ν n
→
the first Galoiscohomology group. Suppose k contains a primitive nth root of
unity. By fixing a primitive nth root of unity, we identify H1(k,Z/nZ) with
k∗/k∗n. With this identification we have ∂ ((a,b) ) = aν(b) k∗/k∗n, where
∗ ν n bν(a)∗ ∈
for any c K which is a unit at ν, c denotes its image in k . More generally,
∈
let E/K be a cyclic unramified inert extension of K and σ Gal(E/K) be a
∗ ∈
generator. Let π K be a parameter at ν. Then the residue of (E/K,σ,π)
∈
is (E /k,σ ), where k is the residue field of K at ν, E is the residue field of
0 0 0
E at the unique extension of ν to E and σ is the image of σ.
0
Let K be a field and n an integer. Then H1(K,Z/nZ) classifies the
equivalence classes of pairs (E,σ), where E is a cyclic Galois field extension
of K of degree a divisor of n and σ a generator of the Galois group G(E/K)
of E/K. Let (E,σ) be a pair representing an element in H1(K,Z/nZ). Let
m 1 be an integer. We now describe m(E,σ) in the group H1(K,Z/nZ).
≥
Let d be the greatest common divisor of m and [E : K]. Let E(m) be the
subfield of E fixed by σ[E:K]/d and σ(m) be the restriction of σm/d to E(m).
Then m(E,σ) = (E(m),σm/d). The order of (E,σ) in H1(K,Z/nZ) is equal
to [E : K]. Since the identity element of H1(K,Z/nZ) is (K,id), we have
m(E,σ) is trivial in H1(K,Z/nZ) if and only if E(m) = K if and only if
[E : K] divides m.
Let K be a complete discretely valued field with residue field k. Let L/K
be a finite extension of K. Since K is complete, the discrete valuation of K
extends uniquely to a discrete valuation of L and L is complete with respect
to this discrete valuation. Let n be a natural number which is coprime to
3
the characteristic of k. Let L/K be a Galois extension of degree n. Let L
1
be the maximal unramified extension of K. Since n is coprime to char(k),
the residue field L of L is same as the residue field of L . Since L/K is
0 1
Galois, L /K and L /k are also Galois and there is a natural isomorphism
1 0
Gal(L/K) Gal(L /k). Let E /k be a cyclic extension of degree n. Then
0 0
→
there exists a unique (up to isomorphism) unramified cyclic extension E of
K with residue field E and a natural isomorphism Gal(E/K) Gal(E /k).
0 0
→
Let (E ,σ ) H1(k,Z/nZ). Then we have a unique (E,σ) H1(K,Z/nZ)
0 0
∈ ∈
with [E : K] = [E : k], E as the residue field of E and the image of σ equal
0 0
to σ ; (E,σ) is called the lift of (E ,σ ).
0 0 0
Let be a regular integral scheme with function field F. Let n be an
X
integer which is a unit on . Let f F and P be a point. If f is regular
X ∈ ∈ X
at P, then we denote its image in the residue field κ(P) at P by f(P). Let
1 denote the set of codimension one points of . For each codimension
X X
one point x of , we have discrete valuation ν on F. Let κ(x) denote the
x
X
residue field at x. Since n is a unit on , n is coprime to char(κ(x)) and
X
we have the residue homomorphism ∂ : Br(F) H1(κ(x),Z/nZ). Let
x n
→
α Br(F). We say that α is unramified at x if ∂ (α) = 0. We say that
n x
∈
α is unramified on if it is unramified at every codimension one point of
X
. Let A be a central simple algebra over F. We say tat A is unramified
X
if its class in Br(F) is unramified. If = Spec(B), then we say that α is
X
unramified on B if it is unramified on .
X
Let F bea field andAa central simple algebra over F. Then A M (D)
m
≃
for some central division algebra D over F. The degree of A is defined as
√dim A and the index of A is defined as the degree of D. If L/K is a field
F
extension, thenindex(A L) divides index(A). Let B beanintegral domain
F
⊗
and F its field of fractions. Let be an Azumaya algebra over B, then we
A
define the index of to be the index of F.
B
A A⊗
Let B be a regular integral domain of dimension at most 2 which is
complete with respect to a prime ideal P. Assume that B/P is a regular
integral domain of dimension at most 1 (for example P is a maximal ideal).
Let κ(P) be the field of fractions of B/P. Let A be a central simple algebra
over the field of fractions F of B which is unramified on B. Then there exists
an Azumaya algebra over B such that F A ([CTS], 6.13)). For
B
A A⊗ ≃
an ideal I of B, we denote the algebra B/I by A(I).
B
A⊗
Lemma 1.1 Let A, B, and P be as above. Then index(A) = index
A
( κ(P)).
B/P
A⊗
4
Proof. Suppose A M (D) for some division algebra D over F. Since
n
≃
A is unramified on B, D is also unramified on B. Since B is a regular
domain of dimension 2, there exists an Azumaya algebra over B such that
D
K D ([CTS], 6.13)). Since Br(B) Br(F) is injective ([AG],
B
D ⊗ ≃ →
7.2), = in Br(B). In particular κ(P) = κ(P) in
B/P B/P
A D A ⊗ D ⊗
Br(κ(P)). Hence index(A) = degree(D) = dim (D) = rank ( ) =
F B
D
rank ( B/P) = dim ( κp(P)) index( p κ(P)).
B/P B κ(P) B/P B/P
D ⊗ D⊗ ≥ A⊗
p Suppose κ(P)p M (D ) for some central division algebra D
B/P m 0 0
A ⊗ ≃
over κ(P). Since B/P is regular integral domain of dimension 1, there exists
an Azumaya algebra over B/P such that κ(P) D . Since
0 0 B/P 0
D D ⊗ ≃
B is P-adically complete, there exists an Azumaya algebra ˜ over B such
0
˜ ˜ D
that B/P and = in Br(B) ([C], [KOS]). In particu-
0 B 0 0
lar AD= ⊗˜ F ≃in DBr(F). AHenceDindex( κ(P)) = degree(D ) =
0 B B/P 0
D ⊗ A ⊗
˜ ˜
dim (D ) = rank ( ) = rank ( ) = dim ( F)
κ(P) 0 B/P 0 B 0 F 0
ipndex(A). Thus indpex(A) = inDdex( q κ(P)D). q D ⊗ ≥(cid:3)
B/P
A⊗
Let K be a field and L a finite extension of K. Then L is called K-
adequate ifthereisadivisionringD centraloverK containingLasamaximal
subfield. A finite group G is called K-admissible if there is a Galoisextension
L of K with G as the Galois group of L over K, and L is K-adequate.
A finite group G is called metacyclic if G has a normal subgroup H such
that H is cyclic and G/H is cyclic.
2. A local-global principle for central simple
algebras
Let K be a complete discretely valued field with residue field k. Let F be
the function field of a curve over K. Let n be an integer which is coprime
to the characteristic of k. Assume that K contains a primitive nth root of
unity. In this section we prove a certain Hasse principle for central simple
algebras over F of index n. We begin with the following
Lemma 2.1. (cf. [FS], Proposition 1(3) and [JW], 5.15) Let R be a com-
plete discrete valuated ring and K its field of fractions. Let A be a central
simple algebra over K of index n which is unramified at R. Let E be an
unramified cyclic extension of K of degree m and σ a generator of the Galois
group of E/K. Let π be a parameter in R. Assume that mn is invertible in
R. Then index(A (E,σ,π)) = index(A E) [E : K].
⊗ ⊗ ·
5
The following two lemmas (2.2, 2.3) are well known.
Lemma 2.2. Let R be a regular ring of dimension 2 with field of fractions
F. Let n be an integer which is a unit in R. Assume that F contains a
primitive nth root of unity. Suppose A is a central simple algebra over F
which is unramified on R. Let x R be a regular prime and ν be the
∈
discrete valuation on F given by x. Suppose that R is complete with respect
to (x)-adic topology. Let u R be a unit. Then index(A F (√n u)) =
ν
∈ ⊗
index(A F(√n u)).
⊗
Proof. Let S be the integral closure of R in F(√n u). Since R is a regular
ring and n, u are units in R, S is also a regular ring. Let x be a regular
prime in R, then x is also a regular prime in S. Thus by replacing R by S,
it is enough to show that index(A F ) = index(A).
ν
⊗
Since R is a two-dimensional regular ring, there is an Azumaya algebra
over R with F A ([CTS], 6.13)). Since R is complete with respect
A A⊗ ≃
to (x)-adic topology, index(A) = index( κ(x)) (cf. 1.1), where κ(x) is
A ⊗
the field of fractions of R/(x). Let R be the ring of integers in F . Since
ν ν
A is unramified on R, A is also unramified on R . Since F is complete,
ν ν
index(A F ) = index( κ(ν)) (cf. 1.1). Since κ(ν) = κ(x), index(A) =
ν
⊗ A ⊗
index(A F ). (cid:3)
ν
⊗
Lemma 2.3. LetRbeacompleteregularlocalringwithfieldoffractionsF
and residue field k. Let n be an integer which is a unit in R. Let u , ,u
1 r
··· ∈
R beunits. Suppose x R is a regular prime. Let ν bethe discrete valuation
∈
on F given by x. Then [F (√n u , , √n u ) : F ] = [F(√n u , , √n u ) : F].
ν 1 r ν 1 r
··· ···
Proof. Since F is a complete discretely valued field and n, u , , u are
ν 1 r
···
unitsintheringofintegers, [F (√n u , , √n u ) : F ] = [κ(ν)(√n u , , √n u ) :
ν 1 r ν 1 r
··· ···
κ(ν)]. Since κ(ν) is the field of fractions of the complete local ring R/(x)
and the the residue field of R/(x) is k, we have [κ(ν)(√n u , , √n u ) :
1 r
···
κ(ν)] = [k(√n u , , √n u ) : k]. Since R is complete and u , ,u are
1 r 1 r
··· ···
units, we also have [F(√n u , , √n u ) : F] = [k(√n u , , √n u ) : k]. Hence
1 r 1 r
··· ···
[F (√n u , , √n u ) : F ] = [F(√n u , , √n u ) : F]. (cid:3)
ν 1 r ν 1 r
··· ···
Proposition 2.4. Let R be a 2-dimensional complete regular local ring
with maximal ideal m = (x,y). Let F be the field of fraction of R and k the
residue field of R. Let n be an integer coprime to char(k). Assume that F
contains a primitive nth root of unity. Let A be a central simple algebra over
F of degree n. Suppose that A is unramified on R except possibly at x and
6
y. Then index(A) = index(A F ) for the discrete valuation ν of F given
ν
⊗
by the prime ideals either (x) or (y) of R.
Proof. Suppose that A is unramified on R. Let ν be the discrete valuation
ofF givenbyx. SinceAisunramifiedonR,by(2.2),wehaveindex(A F ) =
ν
⊗
index(A).
Suppose that A is ramified on R only at the prime ideal (x). Let ν be
the discrete valuation on F given by the prime ideal (x) of R. By ([S1]), we
′ ′
have A = A (u,x) for some unit u in R and a central simple algebra A
⊗
over F which is unramified on R. Since F is complete and u is a unit at
ν
ν, there is a unique extension of ν to F (√n u) such that F (√n u) is complete
ν ν
with the residue field κ(ν)(√n u), where u is the image of u in κ(ν). Since A′
is unramified on R, we have
′
index(A F ) = index(A (u,x) F )
ν ν
⊗ = index(A′ ⊗F (√n u⊗)) [F (√n u) : F ] (by (2.1))
ν ν ν
= index(A′ ⊗F(√n u)) ·[F (√n u) : F ] (by (2.2))
ν ν
= index(A′ ⊗F(√n u))·[F(√n u) : F] (by (2.3)).
⊗ ·
Since A = A′ (u,x), the index of A divides index(A′ F(√n u)) [F(√n u) :
⊗ ⊗ ·
F] = index(A F ). Thus index(A) = index(A F ).
ν ν
⊗ ⊗
Assume that A is ramified on R at both the primes (x) and (y). Then by
([S1]), either A = A′ (u ,x) (u ,y) or A = A′ (uyr,x) where u ,u ,u
1 2 1 2
⊗ ⊗ ′ ⊗
are units in R, r coprime with n and A unramified on R.
′ ′
Suppose that A = A (u ,x) (u ,y) for some units u ,u R and A
1 2 1 2
⊗ ⊗ ∈
unramified on R. Let ν be the discrete valuation on F given by y. Then by
′
(2.1), we have index(A F ) = index(A (u ,x) F (√n u )) [F (√n u ) :
ν 1 ν 2 ν 2
⊗ ⊗ ⊗ ·
F ]. By (2.2), we have [F (√n u ) : F ] = [F(√n u ) : F]. We now compute
ν ν 2 ν 2
′
index(A (u ,x) F (√n u )).
1 ν 2
⊗′ ⊗ ′
Since A (u ,x) is unramified at ν and u is a unit at ν, index(A
1 2
(u ,x) F (√n⊗u )) = index(A′ (u ,x) κ(ν)(√n u ))(cf. 1.1), whereu isth⊗e
1 ν 2 1 2 2
⊗ ⊗ ⊗
image of u in κ(ν). Since κ(ν) is the field of fractions of R/(y), the image of
2
x in κ(ν) gives a discrete valuation µ on the residue field κ(ν) with κ(µ) = k
and κ(ν) is complete with respect to µ. Since u is a unit at µ, κ(ν)(√n u )
2 2
is a complete discrete valuated field with x as parameter and residue field
k(√n u ). Thus, we have
2
index(A′ (u ,x) F (√n u )) = index(A′ (u ,x) κ(ν)(√n u ))(by (1.1))
1 ν 2 1 2
= index(A⊗′ κ(ν)(⊗√n u , √n u )) [κ(ν)(√n u⊗, √n u ) : κ⊗(ν)(√n u )] (by (2.1))
2 1 2 1 2
′ ⊗ ·
= index(A F (√n u , √n u )) [F (√n u , √n u ) : F (√n u )] (by (1.1))
ν 2 1 ν 2 1 ν 2
′ ⊗ ·
= index(A F(√n u , √n u )) [F(√n u , √n u ) : F(√n u )] (by (2.2),(2.3)).
2 1 2 1 2
⊗ ·
7
Hence
′
index(A F ) = index(A (u ,x) F (√n u )) [F (√n u ) : F ](by (2.1))
ν 1 ν 2 ν 2 ν
⊗′ ⊗ ⊗ ·
= index(A F(√n u , √n u )) [F(√n u , √n u ) : F(√n u )] [F(√n u ) : F]
2 1 2 1 2 2
′ ⊗ · ·
= index(A F(√n u , √n u )) [F(√n u , √n u ) : F].
2 1 2 1
⊗ ·
′
On the other hand we have index(A) = index(A (u ,x) (u ,y)) divides
1 2
′ ⊗ ⊗
index(A F(√n u , √n u )) [F(√n u , √n u ) : F]. In particular index(A F ) =
1 2 1 2 ν
⊗ · ⊗
index(A).
Assume that A = A′ (uyr,x) for some unit u R, r coprime to n and
′ ⊗ ∈
A unramified on R. Let ν be the discrete valuation on F given by the prime
ideal (x) of R. By (2.1), we have index(A F ) = index(A′ F (√n uyr))
ν ν
⊗ ⊗ ·
[F (√n uyr) : F ]. Since y is a regular prime in R and r is coprime to n, it
ν ν
follows that [F(√n uyr) : F] = [F (√n uyr) : F ] = n. Since F is a complete
ν ν ν
discrete valuated field and A′ is unramified, we have index(A′ F (√n uyr)) =
ν
⊗
index(A′ κ(ν)( n uyr)) (cf. 1.1). Since κ(ν) is a complete discrete valuated
⊗
field with y asa pparameter andresidue field k, κ(ν)( n uyr) is also a complete
′
discrete valuated field with residue field k. Since Apis unramified on R, we
haveindex(A′ κ(ν)( n uyr)) = index(A′ k)(cf. (1.1)). SinceRiscomplete,
⊗ ⊗
by (1.1), we have indpex(A′) = index(A′ k) = index(A′ κ(ν)( n uyr)) =
index(A′ F (√n uyr)). Therefore, we hav⊗e ⊗ p
ν
⊗
index(A F ) = index(A′ (uyr,x) F )
ν ν
⊗ = index(A′ ⊗F (√n uy⊗r)) [F (√n uyr) : F ]
ν ν ν
′ ⊗ ·
= index(A) n.
·
Since A = A′ (uyr,x), index(A) index(A′) n = index(A F ). Thus
ν
⊗ ≤ · ⊗
index(A) = index(A F ). (cid:3)
ν
⊗
Proposition 2.5. Let R be a 2-dimensional regular ring with field of frac-
tion F and n an integer which is a unit in R. Assume that F contains a
primitive nth root of unity. Suppose that R is complete with respect to (s)-
adic topology for some prime element s R with R/(s) a Dedekind domain.
∈
Let ν be the discrete valuation on F given by s. Let A be a central simple
algebra over F of degree n which is unramified on R except at ν. Further
assume that the residue of A at (s) is given by a unit a in R/(s). Then
index(A) = index(A F ).
ν
⊗
Proof. Supposethat Aisunramified onR. Since Ris(s)-adicallycomplete
and R/(s) is a regular domain of dimension 1 with field of fractions κ(ν),
index(A) = index(A κ(ν)) = index(A F ) (cf. (1.1)).
R/(s) ν
⊗ ⊗
8
Suppose that A is ramified on R. Then by the assumption on A, A is
ramified on R only at the prime ideal (s) of R and the residue of A at (s)
is given by a unit a in R/(s). Let u R with image a R/(s). Since R
∈ ∈
is (s)-adically complete and the image a of u modulo (s) is a unit, u is a
unit in R. The cyclic algebra (u,s) is ramified on R only at (s) and the
residue of (u,s) at (s) is (a). Since A is ramified only at the prime ideal
′
(s) of R, and (a) is the residue of A at s, we have A = A (u,s) for some
′ ⊗
central simple algebra A over F which is unramified on R. By (2.1), we
have index(A F ) = index(A′ F (√n u)) [F (√n u) : F ]. Since R is (s)-
ν ν ν ν
⊗ ⊗ ·
adically complete and R/(s) is integrally closed domain with u a unit, we
have [F(√n u) : F] = [F (√n u) : F ]. Since A′ is unramified on R and R is
ν ν
(s)-adically complete, by (2.2), we have index(A′ F (√n u)) = index(A′
ν
F(√n u)). Hence we have index(A F ) = index(A′⊗ F(√n u)) [F(√n u) : F⊗].
ν
⊗ ⊗ ·
In particular index(A F ) = index(A). (cid:3)
ν
⊗
Theorem 2.6. Let T be a complete discrete valuation ring with fraction
field K and residue field k. Let X be a regular, projective, geometrically
integral curve over K and F = K(X) the function field of X. Let l be prime
not equal to char(k) and A a central simple algebra over F of index n = ld
for some d 1. Assume that K contains a primitive nth root of unity. Then
≥
index(A) = index(A F ) for some discrete valuation ν of F.
ν
⊗
Proof. Let A be a central simple algebra over F. We choose a regular
proper model /T of X/K such that the support of the ramification divisor
X
A and the components of the special fibre of /T are a union of regular
X
curves with normal crossings. Let Y = k denote the special fibre.
T
X ×
For each irreducible curve C in the support of ramification divisor of A,
let a κ(C)∗ besuch thatthe residue ofAatC is(a ) H1(κ(C)∗,Z/nZ).
C C
∈ ∈
Let S be a finite set of closed points of the special fibre containing all
singular points of Y and all those points of irreducible curves C where a is
C
not a unit.
For each P S, let F be the field of fractions of the completion Rˆ of
P P
∈
the local ring R of at P. Let t be a uniformizing parameter for T. For
P
X
each irreducible component U of Y S, let R be the ring of elements in F
U
which are regular on U. It is a regula\r ring. Let Rˆ be the completion of R
U U
at (t) and F the field of fractions of Rˆ . As in ([CTPS], proof of 3.1), we
U U
choose S such that for every irreducible component U of Y S, t = u.sr for
some integer r 1, a unit u R and R /s = Rˆ /s is a D\edekind domain
U U U
with field of frac≥tions k(U). In∈particular, the t-adic completion Rˆ coincides
U
with the s-adic completion of R .
U
9
Let be the set of irreducible components of X S. By ([HHK1], 5.1),
U \
we have index(A) = lcm of index(A F ), ζ S. Since index(A F ) is
ζ ζ
⊗ ∈ U ∪ ⊗
a power of the prime l for all ζ P, we have index(A) = index(A F )
ζ
∈ P ∪ ⊗
for some ζ S. In particular index(A) equal to either index(A F ) for
U
∈ U ∪ ⊗
some irreducible components U of Y S or index(A F ) for some P S.
P
\ ⊗ ∈
Suppose that index(A) = index(A F ) for some P S. The local
P
ring Rˆ is a complete regular local ring⊗of dimension 2 wit∈h maximal ideal
P
ˆ
(x,y) such that A is ramified on R at most at x and y. By (2.4), we have
P
index(A) = index(A F ) for the discrete valuation of F given by either
Pν P
⊗
(x) or (y). Since the restriction ν of ν to F is non-trivial, ν is a discrete
0 0
valuation on F and index(A) = index(A F ).
⊗ ν0
Suppose that index(A) = index(A F ) for some irreducible component
U
U of X S. Then, by the choice of U⊗, A is unramified on Rˆ except at (s)
U
\
and the residue at (s) is given by a unit in R /(s). Let ν be the discrete
U
valuation on F given by (s). Since Rˆ is (s)-adically complete, by (2.5),
U U
index(A) = index(A F ). Since the restriction of ν to F is also given
Uν
⊗
by the ideal (s) in R , ν is non-trivial on F. In particular F F and
U ν Uν
index(A) = index(A F ). ⊂ (cid:3).
ν
⊗
Remark 2.7. Let F and A be as in the above theorem. Then by (2.6), it
follows that there exists a discrete valuation ν of F such that index(A) =
index(A F ). From the proof of (2.6) it follows that this discrete valuation
ν
⊗
comes from a codimension one point of a regular proper model of F. In
particular, the residue field κ(ν) is either a finite extension of K or a function
field of a curve over a finite extension of k.
3. Necessary conditions for Admissibility
In this section, we give a necessary condition for a finite group to be admis-
sible over function fields of curves over complete discretely valued fields.
Let K be a complete discretely valued field with residue field k and π
a parameter. Let D be a central division algebra over K of degree n and
L a maximal subfield of D. Suppose that n is coprime to char(k). Let
(E ,σ ) H1(k,Z/nZ) be the residue of D and (E,σ) H1(K,Z/nZ) the
0 0
∈ ∈
lift of (E ,σ ) (cf. 1). We fix an algebraic closure K of K and also an
0 0
§
algebraic closure k of k. All the finite extensions of K and k are considered
as subfields of K and k respectively. Let F = L E. Let L be the maximal
1
∩
unramified extension of F contained in L. Since E/K is unramified, L E =
∩
L E = F.
1
∩
10